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Having picked up a rudimentary understanding of tensors from reading mechanics papers and Wikipedia, I tend to think of rank-2 tensors simply as square matrices (along with appropriate transformation rules). Certainly, if the distinction between vectors and dual vectors is ignored, a rank 2 tensor $T$ seems to be simply a multilinear map $V \times V \rightarrow \mathbb{R}$, and (I think) any such map can be represented by a matrix $\mathbf{A}$ using the mapping $(\mathbf{v},\mathbf{w}) \mapsto \mathbf{v}^T\mathbf{Aw}$.

My question is this: Is this a reasonable way of thinking about things, at least as long as you're working in $\mathbb{R}^n$? Are there any obvious problems or subtle misunderstandings that this naive approach can cause? Does it break down when you deal with something other than $\mathbb{R}^n$? In short, is it "morally wrong"?

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btw, note that there is some inconsistency in terminology: one can speak about (m,n)-tensors — that is, multilinear maps $V^{n}\times (V^*)^m\to\mathbb{R}$ (so your bilinear map is a (0,2)-tensor); and "rank-k tensor" may mean (k,0)-tensor ("covariant") or (0,k)-tensor ("contravariant") or sometimes any (n,m)-tensor with n+m=k. – Grigory M Aug 13 '10 at 21:05
@Kaestur Hakarl: I am not sure what you mean. The correspondence is exact, as one can see from Grigory's answer using an identification of V with its dual. – Qiaochu Yuan Aug 13 '10 at 21:17
@Qiaochu: Yes, you are right. That certainly explains why I couldn't think of anything. – Larry Wang Aug 13 '10 at 23:43
Related Phys.SE question: and links therein. – Qmechanic May 1 '14 at 17:36
@Franck: What is the point of the tensor-rank tag? – Rahul May 16 '14 at 22:42
up vote 10 down vote accepted

You're absolutely right.

Maybe someone will find useful a couple of remarks telling the same story in coordinate-free way:

  • What happens here is indeed identification of space with its dual: so a bilinear map $T\colon V\times V\to\mathbb{R}$ is rewritten as $V\times V^*\to\mathbb{R}$ — which is exactly the same thing as a linear operator $A\colon V\to V$;

  • An identification of $V$ and $V^*$ is exactly the same thing as a scalar product on $V$, and using this scalar product one can write $T(v,w)=(v,Aw)$;

  • So orthogonal change of basis preserves this identification — in terms of Qiaochu Yuan's answer one can see this from the fact that for orthogonal matrix $B^T=B^{-1}$ (moral of the story: if you have a canonical scalar product, there is no difference between $T$ and $A$ whatsoever; and if you don't have one — see Qiaochu Yuan's answer.)

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It's not misleading as long as you change your notion of equivalence. When a matrix represents a linear transformation $V \to V$, the correct notion of equivalence is similarity: $M \simeq B^{-1} MB$ where $B$ is invertible. When a matrix represents a bilinear form $V \times V \to \mathbb{R}$, the correct notion of equivalence is congruence: $M \simeq B^TMB$ where $B$ is invertible. As long as you keep this distinction in mind, you're fine.

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Thanks! I wish I could accept this answer too. It explains why I never encountered this distinction in practice -- I've only been dealing with orthogonal changes of basis. – Rahul Aug 13 '10 at 23:17

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